技術標籤：演算法題連結串列java演算法

題目描述

假設連結串列中每一個節點的值都在 0 - 9 之間，那麼連結串列整體就可以代表一個整數。

給定兩個這種連結串列，請生成代表兩個整數相加值的結果連結串列。

連結串列 1 為 9->3->7，連結串列 2 為 6->3，最後生成新的結果連結串列為 1->0->0->0。

示例

輸入：(6 -> 1 -> 7) + (2 -> 9 -> 5)，即617 + 295
輸出：9 -> 1 -> 2，即912

解決方法

1. 反轉兩個連結串列
2. 同時遍歷兩個連結串列，依次相加生成新節點，並記錄進位
3. 反轉新連結串列

==> 還可以簡單進階一下，不完全生成一條全新的連結串列。也就是說在任選一條連結串列基礎上賦值結果，若判斷少了節點，再新建，並直接連線在該連結串列後

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head1 ListNode類 
     * @param head2 ListNode類 
     * @return ListNode類
     */
 

    public ListNode addInList (ListNode head1, ListNode head2) {
        if (head1 == null || head2 == null) {
            return head1 == null ? head2 : head1;
        }
        head1 = reverse(head1);
        head2 = reverse(head2);
        ListNode ptail = new ListNode(-1);
        ListNode phead =
 
 ptail, pnew = ptail;
        int temp = 0;            //進位
        while (head1 != null || head2 != null || temp > 0) {
            int num1 = (head1 == null ? 0 : head1.val);
            int num2 = (head2 == null ? 0 : head2.val);
            int num = num1 + num2 + temp;
            pnew = new ListNode(num%10);
            ptail.next = pnew;
            ptail = ptail.next;
            temp = num/10;
            head1 = (head1 == null ? null : head1.next);
            head2 = (head2 == null ? null : head2.next);
        }
        phead = reverse(phead.next);
        return phead;
    }
    
    public ListNode reverse (ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode preNode = head;
        ListNode target = head.next;
        while (target != null) {
            preNode.next = target.next;
            target.next = head;
            head = target;
            target = preNode.next;
        }
        return head;
    }
}

進階

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head1 ListNode類 
     * @param head2 ListNode類 
     * @return ListNode類
     */
    public ListNode addInList (ListNode head1, ListNode head2) {
        if (head1 == null || head2 == null) {
            return head1 == null ? head2 : head1;
        }
        ListNode ptail = head2;            //定位尾結點
        head1 = reverse(head1);
        head2 = reverse(head2);
        ListNode phead = head2;
        int temp = 0;            //進位
        while (head1 != null || head2 != null || temp > 0) {
            int num1 = (head1 == null ? 0 : head1.val);
            int num2 = (head2 == null ? 0 : head2.val);
            int num = num1 + num2 + temp;
            if (head2 != null) {
                head2.val = num%10;
            } else {
                ptail.next = new ListNode(num%10);
                ptail = ptail.next;
            }
            temp = num/10;
            head1 = (head1 == null ? null : head1.next);
            head2 = (head2 == null ? null : head2.next);
        }
        phead = reverse(phead);
        return phead;
    }
    
    public ListNode reverse (ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode preNode = head;
        ListNode target = head.next;
        while (target != null) {
            preNode.next = target.next;
            target.next = head;
            head = target;
            target = preNode.next;
        }
        return head;
    }
}